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3 years ago

11

Mia is drawing an angle that is made up of 35 one-degree angles. How many degrees is Mia's angle?

1 answer:

Tcecarenko [31]3 years ago

3 0

Answer:

35 degrees

Step-by-step explanation:

add all the one degrees together

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Let X be a continuous random variable with density functionf(x)={1−|x|0 for −1

BlackZzzverrR [31]

Answer:

$f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1$

Step-by-step explanation:

For this case we want to find the density function for $Y=X^2$

And we have the following density function for the random variable X:

$f(X) = 1- |X|,-1 \leq X \leq 1$

So we can do this replacing $Y=X^2$

$F_Y (Y \leq y) = P(X^2 \leq y)$

If we apply square root on both sides we got:

$P(-\sqrt{y} \leq X \leq \sqrt{y}) = \int_{-\sqrt{y}}^0 1+t dt +\int_{0}^{\sqrt{y}} 1-t dt$

And if we integrate we got this:

$F_Y (y) = [t+ \frac{t^2}{2}] \Big|_{-\sqrt{y}}^0+ [t -\frac{t^2}{2}] \Big|_{0}^{\sqrt{y}}$

And replacing we got:

$F_Y (y) = [0 -(-\sqrt{y} +\frac{y}{2})] + [\sqrt{y} -\frac{y}{2}]$

$F_Y (y) = 2 |\sqrt{y}| -y$

And if we want to find the density function we just need to derivate the pdf like this:

$f_Y (y) =\frac{d}{dy} = 2 \frac{1}{2 |\sqrt{y}|} -1 = \frac{1}{|\sqrt{y}|} -1,0 \leq Y\leq 1$

3 0

4 years ago

Factor completely. 5x^2-20x+20=5x 2 −20x+20

Harman [31]

Step-by-step explanation:

5x^2 - 20x + 20 = 5x^2 - 20x + 20

5(x^2 - 4x + 4) = 5(x^2 - 4x + 4)

(5 + 5) (x^2 - 4x + 4)

5 0

3 years ago

The Environmental Protection Agency records data on the fuel economy of many different makes of cars. Data on the mileage of 20

artcher [175]

Answer:

A) 8.5 miles per gallon

Step-by-step explanation:

Given

$x = \{12\ 13\ 15\ 16\ 16\ 17\ 18\ 18\ 19\ 19\ 20\ 20\ 22\ 23\ 24\ 26\ 26\ 27\ 27\ 29\}$

$n = 20$

Required

The IQR

This is calculated as:

$IQR = Q_3 - Q_1$

Calculate $Q_1$

$Q_1 = \frac{1}{4}(n + 1)th$

$Q_1 = \frac{1}{4}(20 + 1)th$

$Q_1 = \frac{1}{4}(21)th$

Remove bracket

$Q_1 = 5.25th$

This means that:

$Q_1$ is the mean of 5th and 6th item

From the given data:

$5th \to 16$

$6th \to 17$

So, we have:

$Q_1 = \frac{1}{2}(16 + 17)$

$Q_1 = \frac{1}{2}(33)$

$Q_1 = 16.5$

$Q_1 = 16.25$

Calculate $Q_3$

$Q_3 = \frac{3}{4}(n + 1)th$

$Q_3 = \frac{3}{4}(20 + 1)th$

$Q_3 = \frac{3}{4}(21)th$

Remove bracket

$Q_3 = 15.75th$

This means that:

$Q_3$ is the mean of 15th and 16th item

From the data:

$15th \to 24$

$16th \to 26$

So:

$Q_3 = \frac{1}{2}(24 + 26)$

$Q_3 = \frac{1}{2}(50)$

$Q_3 = 25$

Recall that:

$IQR = Q_3 - Q_1$

$IQR = 25 -16.5$

$IQR = 8.5$

7 0

3 years ago

Maths question pls answer

Vitek1552 [10]

Answer:

81

Step-by-step explanation:

We need to know the sides of the square so 36÷4=9

Area of a square = Side x side

Area of square = 9x9=81

3 0

3 years ago

Read 2 more answers

If 4 friends share a bag of potato chips that costs $4.20, how much should each pay? a. $0.20 c. $10.50 b. $0.11 d. $1.05

IrinaVladis [17]

Answer:

D $1.05

Step-by-step explanation:

$4.20 divided by the amount of friends (4) is equal to $1.05 per friend.

5 0

3 years ago

Read 2 more answers